Question 362857
The point is that there may be 2 solution sets in this type of questions..

First >> if  (x^2-2x)=>0 >>  x^2-2x = 3x-6 >> x^2-5x+6=0 >> (x-2)*(x-3)=0 >> x={2,3} 

Second >> (x^2-2x)<0 >> x^2-2x = -3x+6 >> x^2+x-6=0 >> (x-2)*(x+3)=0 >> x={2,-3}

RF.